 Welcome back everyone, we are going to continue our discussion from the last class in which we saw basically we learn about Duhamel integral and we also saw that how to obtain solution of a single degree of freedom system subject to any arbitrary excitation ok. Now we are going to discuss some of the specific cases of pulse excitation as well as step forces and then see whether how we can obtain the analytical response of a single degree of freedom system. Last class we saw that if we have some arbitrary excitation ok, so if we have the equation a single degree of freedom system which is represented by let us say this differential equation ok and I am not considering damping here for the time being and if P t is some arbitrary excitation ok. So, we derived a integral which is called a convolution integral ok to find out the response to any arbitrary excitation P of t and the idea behind that was that if I have let us say any random ok any arbitrary varying load or excitation P t then the response at any time t ok to get this response we divided this force into very small duration forces ok. And we said that these forces small duration forces would basically represent a impulse ok. So, they would represent each of them an impulse and their response at time t would be some of responses due to each of impulses ok and as we know the response due to an impulse is basically free vibration with the initial velocity ok and we sum up all those responses up to time t to get the response u t at time t. So, basically we derived the expression first for Duhamel convolution integral and we saw that we can substitute the value of unit impulse response function to get the Duhamel integral ok. So, I am just going to write the final expression ok. So, for an undamped system we got this expression ok. So, we can utilize this expression to get the response ok and similarly for damped system as well we can write down the Duhamel integral. But now I would have omega t here in this there would be a e to the power minus eta this term here times sin omega t t minus tau d tau ok. And we discussed that in some cases Duhamel integral is very useful to get the response ok analytical response of function u t something like this ok and in some cases you can use the conventional method of solving the differential equation and it will purely depends on the problem ok. So, we will have to inspect the problem and look at the integrand that you have inside that integration to see which would be easier ok. So, utilizing this basically we obtain the response to a step force which is represented something like this ok. So, force sorry there is no ok. So, this is a step force ok which suddenly applied low of load of magnitude p naught and then it is maintained over time as p naught ok. And we found out the response to this ok as u t equal to u s t naught which was p naught by k times 1 minus cos omega n t. And then we found out what is the response due to a linearly increasing force which this say this is p t here and this is t and let us say this force is represented by p naught t by t r where p naught is basically this amplitude which is achieved over time duration of t r ok. So, during the linearly increasing linearly increasing phase we try to obtain the solution u t as u s t naught ok times t by t r times sin omega n ok times t divided by omega n t r. So, as you can see it is vibrating at frequency omega n around its static solution which is represented by this here. So, u s t is represented by this p naught t by k t r which is basically this function here ok. So, this is. So, we saw that t r by t n is a very important parameter here and we are further going to discuss that today ok. So, let us now discuss we stopped at this point and let us now discuss what happens to the response during the constant force phase ok. So, what we are going to do today is basically a step force ok with finite rise time all right and this is a better representation of force that is applied in real life. So, if you think about it is almost impossible to apply a sudden force there is always some duration however small it may be ok over which this p naught or the amplitude of the force that is achieved ok. So, in general I can represent any force that reaches to its maximum value p naught through a step force with a finite rise time and we can also call it a ramp force ok. So, basically the problem statement that we have here is basically I am applying a force which goes to its amplitude p naught over a time duration t r ok and then it is maintained at that value p naught ok. So, if I want to find out the response to this step force let us see how to do that ok. So, our differential equation would be mu dot plus k u that is equal to p t ok and this is basically equal to p naught t by t r for t smaller than t r ok. So, for this zone here ok I am representing this function ok all right and for a constant force phase of course it would be p naught for t greater than t r. So, for this we have already obtained the solution and let me write that again ok I have obtained the solution as u s t naught times t by t r minus sin omega n t r divided by omega n t r ok and this is for t smaller than t r. Now, let us obtain solution for the second part now if you look at it ok the second part we have something like this ok and we know this is nothing but the differential equation for a step force that we had done previous to this step force with the finite rise time. So, basically what I am trying to do here the total response at any time after the rise time t r would be ok I am going to divide this force into 2. So, this and this here ok. So, let us say and this is p naught ok. So, up to this point my single degree of freedom system would have achieved some displacement and velocity ok. So, if you represent this instead of p t if you say let us say this is u t it would have achieved some u t here and some u dot ok. Let us say not u t, let us say this is u t r and then u dot t r ok and for a linear system if I am representing the total solution ok to this step function as a sum of solution to this linearly increasing phase up to time t r and then step force from point t r onwards. It would be basically with this any initial condition it would undergo free vibration right that we already know. However, note that the motion starts here at t r instead of t 0 ok and in this case we already know what is the solution for this one is it it we know that ok for this step force the solution is basically ok. So, this is step force and the solution I can write it as u s t naught 1 minus cos omega n t ok and again remember that this is starting at t equal to t r. So, this t I will have to shift to t r ok. So, if I sum it up basically what I have here the total response can be represented as free vibration response due to linearly increasing phase ok and because it is starting at t equal to t r and not t equal to 0 I am basically shifting it by t r. So, that by time ok variable is represented by now t minus t r ok and this I will write it like that sin omega n minus t minus t r ok plus there is a constant force phase ok for which the response I can again write it as u s t naught ok 1 minus cos omega n t minus t r all right. Now, this this is the free vibration phase response due to free vibration and this is the constant force phase. So, this is the response due to free vibration this is the constant force phase. Now, in this I can substitute the value of u of t r and u dot of t r by substituting t equal to t r in this expression first and then differentiating it and then again substituting t equal to t r ok. So, I will do that what I am going to do here just write the final expression ok. So, that my u of t is equal to u s t naught ok times 1 plus omega n t r times 1 minus cos omega n t r ok times sin omega n t minus t r and then I have sin omega n t r times cos omega n t minus t r ok. So, this is the expression that I have and if you look at it this is nothing this is a constant here ok you can call it a and this is another constant is b ok and these are coefficients to sin omega n t minus t r and cos omega n minus t r. So, this is basically something of form a cos theta and then b sin theta ok and we will utilize this property later when we have to find out the maximum value of u t ok. But right now let us further simplify this so that my u t becomes u s t naught and the terms that I have inside the bracket ok it reduces to omega n t r and then I have these terms here omega n t r minus cos omega n t minus t r ok alright. So, if you look at it here ok the I can write down my u t divided by u s t naught as this function that I have inside this one ok sin omega n and I can do may be one more thing remember omega n t r I can write it as 2 pi t r by t n ok. So, let me substitute that here so that I get this one as ok t n divided by 2 pi t r and then inside I would have the sin term 2 pi t r by t n ok then this cos term here which would be cos 2 pi t by t minus 2 pi t r by t n ok alright. So, if you look at this carefully ok my system is vibrating at frequency as frequency omega n or the time period t n alright. So, that is cleared vibrates at this natural frequency ok and my response actually depends on the value of t r divided by t n note that it does not individually depends on t r or t n ok. If you look at it it is always in the ratio form anywhere in this expression here for u t divided by u s t naught ok. So, it depends on the ratio. So, I might have different ratio of t r by t r for example, 1 divided by 2 or 2 divided by 4 as long as the ratio is still the same I would get similar kind of dynamic response for my system ok. So, that is an important point that the response actually depends on the value of t r which is the rise time divided by the time period of the system ok. So, that need to be kept in mind ok. So, if we want to plot this for different value of t r by t n ok I will have ok the response of the system ok. So, let me just copy that and then show you how does it look like ok. So, if you look at carefully here ok. So, what these plots basically represent the ratio of the dynamic displacement versus peak static displacement ok and these have been plotted for different values of t r by t n ok and the horizontal axis is basically the normalized time ok. So, if you look at it then my t r is very small like this ok in this case. Note that there are two lines here the dotted line is actually it is static displacement which is nothing but u static of time variation of the static displacement as p t divided by k ok. So, this is the static displacement assuming my system does not have any mass so that I can write it as this and which for this case like you know is simply this dotted line that has been shown here ok. The dynamic displacement is basically shown here using the solid line ok. So, the solid line is my dynamic displacement. Now, let us start with the smallest value of t r by t n ok. So, let me draw a representation of this one. When t r by t n is something like this ok it is starting from 0, but the rise time is very small ok. The response that I see it is similar to a step force and that should be obvious considering ok for very small value of t r by t n ok. The force actually resembles a step force. So, this is p naught by k displacement and then my system basically vibrates around the static equilibrium position which is around the point p naught by k ok. So, I can see here that my solution or my system actually vibrates around the static equilibrium position ok and the vibration, the amplitude, the amplification or let us say difference between the dynamic and the static it is quite large ok. Now, as I increase that as I go from t r by t n to this value that I have here ok. So, as I go from t r by t n to t r by t n equal to 0.5, I see that there is still lot of vibration. So, there is still like you know significant difference between static and dynamic. However, in this case that difference has actually been reduced ok compared to t r by t n equal to 0.2. So, as I increase my ok as I increase the rise time ok, the difference between the static displacement and the dynamic peak displacement actually reduces ok and that is further enforced by the fact when you look at t r by t n equal to response for t r by t n equal to 1.5 and then further response for t r by t n equal to 2.5 ok. For the value very large like t r by t n equal to 2.5 you can see the dynamic solution almost trailing the static solution. So, it is almost like you know close to very close to the static solution. So, that means if I have rise time which is ok which is very very large ok which is very very large then my dynamic solution is almost close to the static solution and I do not see much difference or I do not see much amplification of the response ok. So, in this case depending upon the t r t r by t n ok the dynamic characteristic of the response is determined. So, t r by t n is a very it is a very important parameter to characterize the response to a step force with finite rise time ok. There are two additional cases when t r by t n is equal to 1 and when t r by t n is equal to 2 or for a matter of case whenever you will have t r by t n equal to n ok where n is equal to 1 2 3 so on integer values what you will see ok at the end of the linearly increasing phase you do not get any oscillation ok. So, no vibration during the constant force phase ok and why is that happens is actually when you have this condition your velocity actually is equal to 0 at time t equal to t r ok. So, for that case the velocity is equal to 0 basically if you substitute in the expression for ut you will get ut is equal to ust naught ok which would be constant and not vibrating with time. And if you want to imagine this physically think it like the equilibrium position under the action of load p naught is actually at a displacement p naught by k. So, if you apply a static load of p naught the equilibrium position is p naught by k and let us say I have a spring which is ok vibrating around this p naught by k. Now, when it comes to this displacement p naught by k and if your velocity is 0 then what do you think will happen? Remember in previous instances of free vibration ok when you had system undergoing through the equilibrium position it had the maximum velocity and 0 displacement measured from the equilibrium position. So, if I measure the displacement from the equilibrium position when it crosses here it will have 0 displacement and it would be in equilibrium because that is the equilibrium under the load p naught. So, if it is in equilibrium and there is no velocity there then there is nothing that would take the system beyond this and then bring it back here ok. So, and that happens for very specific case when t r by t n is equal to any integer ok. So, for those cases when the system passes through equilibrium basically it has 0 velocity and then there is nothing that would take the system further on to continue the vibration ok. So, that is why for these cases you get this kind of response no vibration during the constant force phase ok alright. If this is clear now let us look at if you remember for a harmonic excitation ok we had defined a function r d which we said is the ratio of maximum peak dynamic displacement divided by the peak static displacement ok. And that was a function of omega by omega n and this was for the harmonic case. Now I want to do something similar for a step force with finite rise time or for the ramp load ok. However, there is no frequency here there is no excitation frequency in this case ok. So, let us see what do we get ok as the value of unit by US t naught ok and then we will try to go more into the expression ok. So, if you remember my ut ok I wrote down an expression for ut which was US t naught ok times 1 plus 1 divided by omega t r and then there was this term here 1 minus cos omega n t r times sin omega n t minus t r and then there was this term here omega n t r ok times cos omega n t minus t r ok. Now if you look at it as I previously stated this is a constant and this is also a constant only this and this ok sin and cos functions are basically functions of time. So, you can say this is A and this is B and this is as sin theta and let us say this is cos theta ok. So, this is something like this A sin theta minus cos theta and again this is also a constant. So, maximum value of the dynamic displacement basically depends on the maximum value of this expression here and as we know if we have a function of the type A sin theta minus or plus cos B cos theta the maximum value is always A square plus B square ok ok because I have this term sin theta minus phi whatever I can write it and the maximum value is actually A square plus B square under root ok. So, I am going to do that ok I am substituted here ok and when I simplify this I will find out that my U naught by U s t naught is actually this expression here ok I can just I am just going to write the final expression ok. So, this is phi sorry this is t r here phi t r divided by t n and this divided by phi t r by t n. So, again if you look at carefully and let us say this is my r d now ok this depends on the value of t r by t n. So, like for a harmonic excitation of a single degree of freedom system the parameter on which the response modification factor ok depends on is omega by omega n for a ramp loading that parameter is t r by t n ok and I can go ahead ok I can go ahead and plot this function ok r d ok function as a value as a function of basically t r by t n. So, for different values of t r by t n I am going to basically plot this function ok now I can do that mathematically or before doing that mathematically let us see for extreme cases what happens I have already told you when t r by t n is very very small ok it is very very small ok for those cases this is almost like a step function ok this is a suddenly applied force ok suddenly applied force ok and for suddenly applied force we saw that my ut was approximately 2 times us t naught ok that means r d was actually equal to 2 and that we saw that from this graph ok if you look at it it was almost close to 2 r d here was almost close to 2 here ok. So, this is one extreme case other extreme cases when t r is much much greater than 1 ok. So, this is this would be like a slowly applied force ok. So, for slowly applied force we saw that ut was almost same as the static displacement. So, the peak basically the sorry this should be here u naught not not ut ok and here also they should be u naught. So, u naught is almost equal to us t naught ok because my dynamic displacement was actually ok very very much comparable to the static displacement throughout the time. So, r d is actually equal to 1 ok. So, for these two extreme cases we know that a very small value of t r by t n but for very large value of t r by t n what is the value of r t ok. Now, we also said that if t r by t n ok equal to some integer ok then for those cases there is no vibration due to constant during the constant force phase ok and if there is no vibration it means that static and dynamic are basically same. So, again for those cases as well u naught the peak dynamic displacement is actually equal to peak static displacements r d is again equal to 1 ok. So, we utilizing these three cases situations or scenarios let us see if we are able to plot r d versus t r by t n ok all right. So, we have let us say we have said that we have said that when t r by t n is very small the value is actually 2. So, it is somewhere around here when it is very large ok then it is close to 1 and we have also said that it is equal to 1 at integer value of t r by t n. So, let us say this is 1, this is 2, this is 3, this is 4 and so on. So, basically the actual response looks like something like this ok with this decreasing with time ok all right. So, depending upon the value of t r by t n ok we can calculate r d and the same thing can also be obtained using you know the expression for r d ok, but we have just tried to obtain this using the analytical or discussion that we just presented. Now, a point here to note that how much would be the r d how much different it would be from the static displacement if this is 1 let us say this is 2 this is basically static. It depends on the value of t r by t n ok and remember these t r by t n decides ok whether it is going to be a suddenly applied force or whether it is going to be a slowly applied force ok. So, whether it would have dynamic effect or whether it would be almost same as static and it would behave like a static ok. Now, this is a very important parameter because most of the loads when it is said that the load that is being applied is let us say 10 kilo Newton ok before this ok before this course you did not care about knowing how the load was applied. So, what was the time variation of the load and if somebody says to you that something is applied slowly it does not simply means that it takes long time to actually apply that load you also need to know ok whether the time how long is the time with respect to the time period of the system ok. So, that is a that is an important parameter. So, I will give you an example ok. So, whether something is slowly applied or whether it is a static force ok and whether some load is like you know suddenly applied and whether it is a dynamic it depends on this overall ratio ok. So, let us say I have two bridges ok one is ok one bridge is basically a rope bridge ok which is very flexible and the second one I have is actually a concrete bridge ok. Now, let us say for the rope bridge my frequency is around 2 second sorry time period is around 2 second and for concrete bridge let us say it is around 0.1 second ok which is typical you know because it is very flexible the rope bridge it would have ok very large time period compared to concrete bridge which has a very small time period of 0.1 second. Now, whether something is slowly applied or whether it is you know suddenly applied depends on this ratio. So, let us say I want to apply a load ok. So, that the dynamic effect is very very less ok. So, let us say I can do that for let us say T r by T n if this ratio is very very high ok then I can say that there is no dynamic amplification ok my RAD would be equal to 1 ok. So, it would behave like static. Now, to achieve the static type condition ok in this case the rise time has to be 2 times 10 20 second and in this case the rise time could be just 2 second alright. So, if you see that here ok for this rope because it is very flexible you need to be applying the load at a much much slower pace or over a very large duration of time to achieve similar kind of response ok compared to concrete bridge in which you can apply the load much quickly and still be able to obtain similar kind of response ok and like you know you can observe this realistically as well. If you jump on a rope bridge it will start oscillating. So, you will see much more dynamic effect ok. So, remember you are applying similar kind of load over a similar duration ok. So, if you jump on a rope bridge or if you jump on a concrete bridge you are applying similar load over a similar duration. However, because the time period of the two bridges are different you would see basically different response ok. So, keep that in mind that whether something is static or dynamic ok it depends on the ratio TR by TM and slow whether something is applied slowly or suddenly ok. It is not only a function of time ok times relative it is actually depends on in a structural dynamics whether something is slow or fast it depends on the ratio of the duration with respect to a time period of the system ok. So, this is something that you need to keep in mind TR by TM which is a very important parameter alright. So, with this we saw that how to get the response of a system subject to a ramp loading or basically a step forward with a finite rise time ok. Now, what we are going to discuss a different type of excitation which is basically called pulse excitations ok. So, we are going to get into pulse excitation ok. Pulse excitation is basically any type of excitation that has a finite duration of application ok and then it is 0 ok. So, you have a load that is applied over a finite duration and then it is 0 before and after that. So, some of the examples would be ok a load that is applied like this ok or a load let us say which is applied like this or load that is applied like this ok or you could have a different type of load even this is ok. So, any type of load that is applied only over a finite duration of time is called a pulse load and as you can imagine if you apply a pulse load then the system does not actually achieved like in a steady state response it would always be vibrating and depending upon whether the system has damping or not it might come to rest for a damped system or might might always keep on vibrating ok with the initial conditions provided by this pulse type motion ok. So, you know there are different type of living several type of pulse type excitation ok. What we are going to study first is basically a very common type of pulse excitation which is rectangular pulse excitation. So, for a rectangular pulse excitation basically I have a rectangular pulse force of magnitude or say the amplitude P naught applied over a time duration of TD. So, now there is a new parameter remember there was a rise time in the previous case that we had discussed now that there is a different parameter it is called TD ok. So, the load rectangular pulse is applied suddenly it is kept constant till the time duration TD and then it is released. So, that force is 0 and the system that we have to find out the response of the system subject to this type of loading ok. So, I can again go ahead and write down the equation of motion as mu double dot plus k u equal to P of t where P of t is basically P naught for t smaller than TD and equal to 0 for t greater than TD ok. And in terms of solution we are going to follow the same procedure that what we have been doing till now ok. So, first let us do the forced vibration phase ok. So, for pulse type excitation we would always divide our response in forced vibration phase ok and then the free vibration phase ok and then we will try to obtain the response. So, this is basically my forced vibration phase and this is my free vibration phase ok. So, for this case let us first obtain the response for forced vibration phase and see what do we get ok. So, for forced vibration the equation that I have is this and we already know the response for ok this type of loading ok. We had already derived this, this is nothing but a step force ok. So, the response for this we have already obtained as ut ok divided by u s t naught this we had obtained as 1 minus cos omega n t which can be further written as 1 minus cos 2 pi t by t n ok. And remember this response is valid only for time is smaller than the TD which is the time duration of the pulse ok. Once this is clear let us obtain the response for free vibration phase ok. So, we want to obtain the response for free vibration phase and for free vibration phase it is nothing but a free vibration with initial conditions as u of TD and u dot of TD ok that was imparted to the system due to the forced vibration ok. So, we are going to utilize these initial conditions and find out the response in the free vibration. So, ut is nothing but u of TD cos omega n. And remember now because the initial conditions are at TD ok I am going to write this equation like this ok. So, this is my expression. So, I can go ahead and substitute these values ok from the expression here ok and simplify it to get the final response ok. So, final response that I get here is actually cos omega n t minus TD minus cos omega n t. Remember this is the response for t greater than TD ok. We can further simplify this and write omega n t equal to 2 pi by sorry 2 pi t by divided by t n ok. So, that would give us this expression here ok 2 pi TD divided by t n times sin 2 pi t by t n minus half TD by t n ok. So, this is the response that we get for t greater than TD ok. And if you look at it this is nothing but this sin function ok. So, let us call this sin theta or whatever and it is multiplied with this coefficient here which is constant ok. So, it does not depend on time it is actually a function of TD by t n ok. So, the normalized response if you look at here ok this is actually UAT by UST0 ok. This system actually vibrates again at its natural frequency ok and it depends on the ratio TD by t n ok. So, like for the case when we had a forced vibration of single degree of freedom system subject to harmonic excitation there the important parameter was omega by omega n for a step function with finite rise time or ramp function the parameter was the rise time divided by the time period ok. For pulse type motions we will see the important parameter is actually TD by t n ok. So, basically my system when applied to this oscillates about its static position ok at time period of oscillation of t n ok. And basically we can plot these functions ok we can plot UT by UST0 ok for different value of TD by t n and I am going to just bring that figure just to explain you how does it look like. So, if you look at here ok depending upon the value of TD by t n all right we have obtained basically different response histories here ok. So, I have obtained UT by UST0 ok and what I have done here basically combined the response for UT by UST0 ok for free for forced vibration phase and the free vibration phase ok and these are the curves that we get ok. And if you look at this carefully depending upon the value of TD by t n ok the maximum can occur during the forced phase ok or during the free vibration phase like in these two cases ok. And this happens when TD by t n actually below the value of 1 by 2 ok like in these cases then the peak occurs after the value of TD ok. Let us get into that ok we have these responses let us let us get into that one thing that we just need to focus here that when we applied the load p naught over the time duration TD my system actually starts vibrating about a shifted equilibrium position which is p naught by k and when I remove the load it come back to its original position which is U equal to 0 and then it vibrates about this position ok. Even though the amplitude is different the same type of behavior is opted for different value of TD by t n except again for these cases where TD by t n is actually an integer when that is the case then what happens at the end of the forced vibration phase we basically get U TD equal to 0 and U dot TD equal to 0 ok when TD by t n is equal to any integer ok 1 2 3 like that ok. So, for those cases ok after the completion of the forced vibration phase we do not see any response ok. So, the system just sits there ok and which can again be interpreted in terms of you know vibration about equilibrium that if you are having a system that is vibrating and it comes to equilibrium and at that point if you have 0 velocity and it is in equilibrium then there would not be any further oscillation because you do not have any velocity or you do not have any initial condition to take the system above that because your force is now 0 ok and if your initial conditions are also 0 then there would not be any subsequent motion. So, that is why for these cases we obtain behavior which is something like this ok. So, just see that how the resistance behave subject to different values of TD by t n. Now, let us come back ok to finding the maximum response or finding the maximum value of this ok. So, finding maximum this which is nothing, but peak dynamic displacement divided by peak static displacement and as previously described this is basically the response modification factor ok and this response modification factor ok. Let us obtain that for free vibration phase and forced vibration phase and overall RAD would be maximum ok of both ok. So, let us first obtain RAD ok. Let us see what is the value of RAD for forced vibration ok. The value of RAD was ok we had the expression 1 minus cos 2 pi t by t n ok. So, this was my expression. So, the maximum value of this function is what it is equal to which occurs at cos 2 pi by t by t n when it is equal to minus 1 then I get the maximum response as this one and for that to occur ok. So, it could be minus 1 minus 2 and so on ok. So, for those that condition let us see what is my t by t n ok. If I substitute it here I would basically get 2 pi t by t n as pi 2 pi and so on. So, my t by t n is actually t n by 2 ok then t n 3 t n by 2 and like that ok. So, basically the maximum response occurs at these durations ok and for at least one maximum to occur ok at least one maximum to occur can you imagine that my t by t n should at least be greater than t n by 2 ok or that scenario let us say this is t d here sorry not t ok. So, just correct that ok. Can you imagine ok it would achieve a value of 2 only if t d by t n is greater than t n by 2 ok. So, the maximum of this function the maximum of r d during forced vibration phase would be 2 only if ok t d by t n is greater than this expression here ok. So, here maybe I should not write t d by t n just write t d here ok that gives me from this expression I have become this ok. So, it has to be greater than half that means I would have at least one peak of function that I have here and let us try to understand that graphically as well. So, I have this function 1 and this function minus cos 2 pi t by t n. So, basically what I am saying this is 1 here ok and then I am going to draw minus cos 2 pi t d by t n. Let us see how does it look like ok this is a function basically which starts from minus 1 ok and it goes to a value of ok basically something like this ok. So, for it to achieve a maximum value if I sum this up it would look like something like this shifted by ok. So, this is the first peak that occurs and that occurs at t n by 2. The second peak occurs at 3 n by 3 t n by 2 and so on ok and if I sum this up this would look like this whole thing would be shifted by one upwards ok. So, it would look like something like this ok this is the response to during the forced vibration phase all right. So, let us say this is u here this is the value is equal to 2 ok and I will say that this is u naught by u s t naught. So, this is 2 and this is t by t n ok or if you t it will write it as t here then it would be t n by 2 3 n by 2. So, I am just writing here as half ok 3 by 2 and so on. So, what I see here this would achieve a value of 2 only if my t d by t n is greater than this value half ok. So, I hope you understand this is the expression that I have for RAD and this expression can achieve a value up to 2 provided this condition is satisfied ok. So, cos 2 by t t by t n at least becomes minus 1 minus 2 and like that ok. For that to happen my t d has to be greater than t n by 2. So, I will write RAD is equal to 2 if my t d is greater than t n by 2. So, the question comes well what happens if t d by ok if t d is smaller than t n by 2 well then the maximum response would be during the force vibration phase whatever the response as at t equal to t d is ok which basically we have cos 2 by t d by t n. So, it would be just same value and you put the value of t d here. So, if you look at here if you substitute t d by t n equal to half then it becomes 2 ok, but the duration always might not be large enough so that you can have at least one peak during the forced vibration phase ok. So, I have obtained two expressions, two conditional expression for the maximum during the forced vibration phase. Let us now go into the free vibration phase. So, for free vibration phase basically we know that ok system undergoes free vibration with initial conditions as ut and ut dot t naught. And I know that if a system is undergoing free vibration with these initial condition the amplitude of vibration can always be obtained as using this expression here ok and this is from your free vibration class ok it is. So, in that case it was u 0 square here that initial condition is at t equal to t d ok. So, this is here and you can substitute the value of u of t d and u dot of t d to obtain the expression for u 0 ok. When you substitute this you basically get this as 2 of u s t naught ok times sin of ok this under root terms is actually square. So, when you take the root it comes as sin of pi t d by t n ok. So, that may already becomes u 0 u t naught this becomes 2 sin pi t by t n ok alright. So, basically what did we see here ok we saw that during the forced vibration phase ok depending upon the value of t d by t n ok the maximum could be 2 or maximum could be this ok. If t d is smaller than t n by 2 then what happens actually ok then what happens in that case my response is still increasing ok. And this response is basically ok. So, this is let us say ut ok this response here is basically your u of t d m then there is some velocity here as well, but the maximum response then occurs during the free vibration phase ok. And this is for the case when let us say t d by t n is here ok this is smaller than half here ok. If this is greater than half then this peak would be inside that ok for that case. So, the response is. So, for that case ok when t d by t n is smaller than half the maximum occurs during the free vibration phase and that maximum is actually this value. So, if we combine both ok what do we actually get ok. So, if we combine the forced vibration phase and the free vibration phase I can get one expression for R d which would be equal to R d equal to U s t naught ok. This should be equal to 2 sin pi t d by t n ok this is for t d by t n is smaller than half ok. So, as I said this is going to govern the peak is going to occur during the free vibration phase if t d by t n is smaller than half. And if it is greater than half then the peak occurs during the forced vibration phase and it is equal to 2 ok. And you can go ahead and plot these two functions ok. So, I am just going to get you the final figure that we have here. Let me just copy this. So, this is basically what I am saying ok. Let me write this here. This is forced vibration remember this is for free responses actually ok free responses for the free response I have 2 of sin pi t d by t n ok. Forced response is basically when it is smaller than half t d by t n is smaller than half this value is basically here which is 1 minus cos pi t d 2 pi t d sorry 2 pi t d by t n and if it is greater than half it is 2. So, if you take the envelope of both these functions you get basically the overall maxima. In this one this is governed by the free response ok and this function is basically 2 sin pi t d by t n and this is basically the value equal to 2 alright. So, if you look at it we are able to obtain the r d as a function of t d by t n ok. And this is similar when we have obtained r d for a harmonic excitation as function of omega by omega n or for a ramp loading we had obtained this as function of t r by t n. In this case we are obtaining as a function of t d which is the load duration divided by the time period of the system ok. And this is the plot of peak response versus some function of t either directly as a function of t n or some multiple of t n it is called a response spectra ok. So, plot of peak response of a single degree of freedom system versus its time period is called response spectra ok and it is a very useful tool that is used by designers in industry. For example, not everybody is going to obtain the differential equation then go through all those calculation and get the peak response ok. If this chart is available ok. So, let us say this chart is given to us for a rectangular pulse loading then say you have a structure of time period t n alright. And you know that a load duration rectangular pulse is being applied with a load duration t d ok. So, that you can find out the t d by t n ok. And what I am saying here that you know t n and you know this load here that is being applied. So, basically p naught and t d is known to you ok and t n is also known to you. So, if those parameters are known to you do not have to solve any differential equation you can find out the maximum value of u naught as r d times u s t naught ok and r d can be directly obtained here. So, let us say for any given value of t d by t n you can just go here and then find out what is the value of t d by t n ok. And if you know u naught then you can also find out the equivalent static force in the structure as k u naught assuming that it is represented as a single degree of freedom system. So, I am saying if you can represent this building as a single degree of freedom system ok something like this with k and m then for that case ok. The equivalent static force is ok k times the peak dining displacement which we can further obtain as k times r d and u s t naught which is basically p naught by the stiffness of the structure. So, the equivalent static force is basically r d times the peak static force ok alright. So, with this the discussion on rectangular pulse duration actually concludes we are going to extend the same procedure to find out the response for other type of pulses in the subsequent classes ok alright. Thank you very much.